Question

To aid in the prevention of tooth decay, it is recommended that drinking water contain 0.800 ppm fluoride (F–).

(a) How many grams of F– must be added to a cylindrical water reservoir having a diameter of 8.96 × 102 m and a depth of 23.16 m?

(b) How many grams of sodium fluoride, NaF, contain this much fluoride?

Answer 1

ppm = 0.8 F-

ppm = 0.8 mg / L

a)

g of F- needed for

D = 8.96*10^2 m

H = 23.16 m

A = PI*D^2/4 = (3.141592)((8.96*10^2)^2)/4 = 630530.08 m^2

V = A*h = 630530.08*23.16 = 14603076.6528 m^3 --> change to L since 1000L = 1 m3

V =14603076652.8 L

then we need

ppm = mg/L

mg = ppm*L = (0.9)(14603076652.8 ) = 13142768987.5 mg of F-

m = 13142768.987 g of F-

b)

g of NaF needed for this

MW of NaF = 41.98871

MW o F = 18.998403

ratio is

18.998403 /41.98871 = 0.45246455535 g of F / total NaF

then

13142768.987 / ratio = 13142768.987 /0.45246455535 = 29047068.6192 g of NaF are needed to contain such mass of F- ions